Hi friends,
I was going through following link

http://www.allaboutcircuits.com/worksheets/switch_c.html

in example 2, two capacitors are placed in parallel, one is charged and another is not, the question at what voltage these two capacitors will get balanced,answer was given, anyway i found out the formula for solving this problem by energy equations and the formula is as below


vc = v(c1/c1+c2)

where vc = voltage acroos capacitor
v = Initial voltage on charged capacitor.

my question is what if i place a resistor between these two capacitors, i tried solving it using KVL, however i couldnt, if anyone has any idea,help will be appreciated.

Thanks

 

Logically thinking the voltage will be the same as without the resistor due to conservation of charge.

 

I do agree with you,however i want to find the equations which will govern the discharging of C1 and charging of C2 at the same time.

 

Use i=C*dV/dt and solve the resultant differential equation.

 

fuzzy_boy,

in example 2, two capacitors are placed in parallel, one is charged and another is not, the question at what voltage these two capacitors will get balanced,answer was given, anyway i found out the formula for solving this problem by energy equations and the formula is as below


vc = v(c1/c1+c2)

where vc = voltage acroos capacitor
v = Initial voltage on charged capacitor.

You did not get that formula from the energy equations. You got it by setting up equations by the conservation of charge.

my question is what if i place a resistor between these two capacitors, i tried solving it using KVL, however i couldnt, if anyone has any idea,help will be appreciated.

You already have resistance between the capacitors from the wires and capacitor plates. Note the following:

Using energy = (1/2)CE^2, we have at the beginning an energy in C1 at 6 volts of 180 microjoules. The energy after connection at 1.875 volts is 17.6 microjoules and 38.7 microjoules for C1 and C2 respectively. What happened to the missing 123.8 microjoules? It was dissipated in the wires and the plate resistance. Notice that if you insert extra resistance between the capacitors above and beyond what the wires have, it will not change the energy dissipated, and conservation of charge will still be maintained. Extra resistance will just make the energy dissipation slower. As mik3 said, if you want the discharge curve, you have to solve a diffy Q with a resistance included in the equation. If you had absolutely no resistance, you would get a infinite spike of current when you connect the capacitors together.

Ratch

 

Energy is measure in Watts? Did I miss a memo? Did a Joule theft take place?

 

thingmaker3,

Energy is measure in Watts? Did I miss a memo? Did a Joule theft take place?

Yes, a silly mistake which I corrected. Thanks.

Ratch